Q176 / 180 · KCET 2025 · Physics
While determining the coefficient of viscosity of the given liquid, a spherical steel ball sinks by a distance $\mathrm{h}=0.9 \mathrm{~m}$. The radius of the ball $\mathrm{r}=\sqrt{3} \times 10^{-3} \mathrm{~m}$. The time taken by the ball to sink in three trails are tabulated as follows.
\begin{array}{|l|l|}
\hline \text { Trial No. } & \text { Time taken by the ball to fall by } \mathrm{h} \text { (in second) } \\
\hline 1 . & 2.75 \\
\hline 2 . & 2.65 \\
\hline 3 . & 2.70 \\
\hline
\end{array}
The difference between the densities of the steel ball and the liquid is $7000 \mathrm{~kg} \mathrm{~m}^{-3}$. If $\mathrm{g}=10 \mathrm{~ms}^{-2}$, then the coefficient of viscosity of the given liquid at room temperature is
- Correct answerA$0.14 \mathrm{~Pa} . \mathrm{s}$
- B$0.14 \times 10^{-3} \mathrm{~Pa} . \mathrm{s}$
- C$14 \mathrm{~Pa} . \mathrm{s}$
- D0.28 Pa.s
Solution
$\begin{aligned} & \mathrm{t}_{\text {avg }}=\frac{2.75+2.65+2.70}{3}=2.7 \mathrm{sec} \\ & \text { Terminal velocity }=\frac{\mathrm{h}}{\mathrm{t}_{\text {avg }}} \\ & \mathrm{V}_{\mathrm{t}}=\frac{1}{3} \mathrm{~m} / \mathrm{s} \\ & \mathrm{n}=\frac{2}{9} \frac{\pi^2 \mathrm{~g}\left(\mathrm{~S}_{\text {stell }}-\mathrm{S}_{\text {liquid }}\right)}{\mathrm{V}_{\mathrm{t}}} \\ & \mathrm{n}=0.14 \mathrm{~Pa} . \mathrm{S}\end{aligned}$
Hence, the correct answer is option: 'A'.